ABC is an equilateral triangle $P$ is a point on $B C$ such that $B P : P C = 2 : 1$ .
Prove that : $9 A P^{2} = 7 A B^{2}$

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Solution

From $Δ A B M$ ,
Let $A M ⊥ B C$
${A B}^{2} = {B M}^{2} + {A M}^{2}$
$= (\frac{3 x}{2})^{2} + {A M}^{2}$
$= \frac{9 x^{2}}{4} + ({A P}^{2} - {M P}^{2})$ ...[from $Δ A M P$ ]
$= \frac{9 x^{2}}{4} + {A P}^{2} - {(\frac{x}{2})}^{2}$
${A B}^{2} = {A P}^{2} + 2 x^{2}$
${A B}^{2} - 2 x^{2} = {A P}^{2}$
$9 {A B}^{2} - 18 x^{2} = 9 {A P}^{2}$ ....[multiplying by 9]
$9 {A B}^{2} - 2 (9 x^{2}) = 9 {A P}^{2}$
$9 {A B}^{2} - 2 (3 x)^{2} = 9 {A P}^{2}$
$9 {A B}^{2} - 2 (A B)^{2} = 9 {A P}^{2}$
$7 {A B}^{2} = 9 {A P}^{2}$
Hence proved.

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